Question:
Roots of equation \(x^3 - 6x + 1 = 0\) lie in the interval
Roots of equation \(x^3 - 6x + 1 = 0\) lie in the interval
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If function changes sign in an interval $\Rightarrow$ at least one root exists there.
Updated On: Apr 16, 2026
- \((2,3)\)
- \((3,4)\)
- \((3,5)\)
- \((4,6)\)
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The Correct Option is A
Solution and Explanation
Concept:
Use Intermediate Value Theorem (sign change method).
Step 1: Define function
\[ f(x) = x^3 - 6x + 1 \]
Step 2: Check at \(x=2\)
\[ f(2) = 8 - 12 + 1 = -3 \]
Step 3: Check at \(x=3\)
\[ f(3) = 27 - 18 + 1 = 10 \]
Step 4: Conclusion
\[ f(2) 0 \Rightarrow \text{root lies in } (2,3) \] Conclusion: \[ (2,3) \]
Step 1: Define function
\[ f(x) = x^3 - 6x + 1 \]
Step 2: Check at \(x=2\)
\[ f(2) = 8 - 12 + 1 = -3 \]
Step 3: Check at \(x=3\)
\[ f(3) = 27 - 18 + 1 = 10 \]
Step 4: Conclusion
\[ f(2) 0 \Rightarrow \text{root lies in } (2,3) \] Conclusion: \[ (2,3) \]
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